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Chapter 17: Problem 56

A uniform \(75-\mathrm{kg}\) cube is attached to a uniform 70 -kg circular shaft as shown, and a couple \(\mathrm{M}\) with a constant magnitude of \(20 \mathrm{N} \cdot \mathrm{m}\) is applied to the shaft. Knowing that \(r=100 \mathrm{mm}\) and \(L=300 \mathrm{mm}\), determine the time required for the angular velocity of the system to increase from \(1000 \mathrm{rpm}\) to \(2000 \mathrm{rpm} .\)

Short Answer

Expert verified
The time required is approximately 7.72 seconds.

Step by step solution

01

Convert RPM to Radians per Second

First, convert the initial and final angular velocities from revolutions per minute (rpm) to radians per second (rad/s). The formula to convert rpm to rad/s is \( \omega = \text{rpm} \times \frac{2\pi}{60} \). Hence, the initial angular velocity \( \omega_i = 1000 \times \frac{2\pi}{60} \approx 104.72 \text{ rad/s} \) and the final angular velocity \( \omega_f = 2000 \times \frac{2\pi}{60} \approx 209.44 \text{ rad/s} \).
02

Calculate Moment of Inertia

The total moment of inertia for the system (cube and shaft) can be calculated by adding the moments of inertia of the cube and the shaft. The cube's moment of inertia about the axis perpendicular to one face is \( I_{cube} = \frac{1}{6} m_{cube} L^2 \). For the shaft, \( I_{shaft} = \frac{1}{2} m_{shaft} r^2 \). Thus, the total moment of inertia is \( I = I_{cube} + I_{shaft} \).
03

Cube's Moment of Inertia

The moment of inertia of the cube, with mass \( m_{cube} = 75 \text{ kg} \) and side length \( L = 0.3 \text{ m} \), is calculated as \( I_{cube} = \frac{1}{6} \times 75 \times (0.3)^2 = 1.125 \text{ kg} \cdot \text{m}^2 \).
04

Shaft's Moment of Inertia

The moment of inertia of the shaft, with mass \( m_{shaft} = 70 \text{ kg} \) and radius \( r = 0.1 \text{ m} \), is calculated as \( I_{shaft} = \frac{1}{2} \times 70 \times (0.1)^2 = 0.35 \text{ kg} \cdot \text{m}^2 \).
05

Total Moment of Inertia

Add the moments of inertia of the cube and the shaft to find the total moment of inertia: \( I = 1.125 + 0.35 = 1.475 \text{ kg} \cdot \text{m}^2 \).
06

Find Angular Acceleration

Use the formula for torque: \( M = I \alpha \), where \( M = 20 \text{ N} \cdot \text{m} \) is the given torque and \( \alpha \) is the angular acceleration. Solving for \( \alpha \), we have \( \alpha = \frac{M}{I} = \frac{20}{1.475} \approx 13.56 \text{ rad/s}^2 \).
07

Determine Time Required

To find the time \( t \) required for the angular velocity to increase from \( \omega_i \) to \( \omega_f \), use the formula \( \omega_f = \omega_i + \alpha t \). Rearranging gives \( t = \frac{\omega_f - \omega_i}{\alpha} = \frac{209.44 - 104.72}{13.56} \approx 7.72 \text{ seconds} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
In rotational motion, the moment of inertia plays a role similar to mass in linear motion. It determines how difficult it is to change an object's rotational state. For an object, it depends on the mass distribution relative to the axis of rotation. The formula varies based on the geometry of the object.
For example:
  • The moment of inertia of a cube about an axis perpendicular to one of its faces is given by \( I_{cube} = \frac{1}{6} m L^2 \), where \( m \) is the mass, and \( L \) is the side length of the cube.
  • For a cylindrical shaft, it's \( I_{shaft} = \frac{1}{2} m r^2 \), where \( r \) is the radius.
By calculating each part separately and summing them, we find the total moment of inertia of a system, crucial for predicting the angular movement of linked rigid bodies.
Torque Calculation
Torque is the rotational analogue of force. It is the measure of how much a force acting on an object causes that object to rotate. The formula is simple: \[ M = I \alpha \]where \( M \) is the torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration. The given exercise involves a consistent torque applied to the shaft, which allows us to calculate angular acceleration.
Understanding torque helps in many fields, like engineering and physics, to understand how different forces will affect rotating objects. In this situation, with a 20 N路m torque applied, we can determine how fast the system spins up or down.
Angular Acceleration
Angular acceleration is how quickly an object's rotational speed changes. It is to rotational motion what acceleration is to linear motion. The relationship with torque and moment of inertia is given by:\[ \alpha = \frac{M}{I} \]where \( \alpha \) is angular acceleration, \( M \) the applied torque, and \( I \) the moment of inertia.
In our example, with a total inertia from both the cube and shaft known, it鈥檚 straightforward to calculate the angular acceleration. Knowing this value helps predict how long it takes for a rotating system to reach a new speed when a specific torque is applied.
Conversion of Units
Conversions are often necessary to get to the correct units for calculations. In this exercise, converting revolution per minute (rpm) to radians per second (rad/s) is essential, as angular velocity in rad/s is compatible with SI units:
  • Use: \( \omega = \text{rpm} \times \frac{2\pi}{60} \)
  • This changes initial \( \omega_i = 1000 \) rpm to approximately \( 104.72 \) rad/s.
  • And, the final \( \omega_f = 2000 \) rpm to approximately \( 209.44 \) rad/s.
Unit conversion ensures each part of the calculation uses a uniform measurement system, reducing errors and ensuring accuracy when solving physics problems.

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Most popular questions from this chapter

A 6 -lb collar \(C\) is attached to a spring and can slide on rod \(A B,\) which in turn can rotate in a horizontal plane. The mass moment of inertia of rod \(A B\) with respect to end \(A\) is 0.35 Ib. \(\mathrm{ft}\). The spring has a constant \(k=15 \mathrm{lb} / \mathrm{in}\), and an undeformed length of \(10 \mathrm{in}\). At the instant shown, the velocity of the collar relative to the rod is zero and the assembly is rotating with an angular velocity of 12 rad/s. Neglecting the effect of friction, determine (a) the angular velocity of the assembly as the collar passes through a point located 7.5 in. from end A of the rod, (b) the corresponding velocity of the collar relative to the rod.

Knowing that the maximum allowable couple that can be applied to a shaft is 15.5 kip?in., determine the maximum horsepower that can be transmitted by the shaft at (a) 180 rpm, (b) 480 rpm.

A uniform slender bar of length \(L=200 \mathrm{mm}\) and mass \(m=0.5 \mathrm{kg}\) is supported by a frictionless horizontal table. Initially the bar is spinning about its mass center \(G\) with a constant angular speed \(\omega_{1}=6 \mathrm{rad} / \mathrm{s}\). Suddenly latch \(D\) is moved to the right and is struck by end \(A\) of the bar. Knowing that the coefficient of restitution between \(A\) and \(D\) is \(e=0.6\), determine the angular velocity of the bar and the velocity of its mass center immediately after the impact.

A semicircular panel with a radius r is attached with hinges to a circular plate with a radius r and initially is held in the vertical position as shown. The plate and the panel are made of the same material and have the same thickness. Knowing that the entire assembly is rotating freely with an initial angular velocity of \(\omega_{0}\), determine the angular velocity of the assembly after the panel has been released and comes to rest against the plate.

The 10 -in-radius brake drum is attached to a larger flywheel which is not shown. The total mass moment of inertia of the flywheel and drum is 16 lb. fit. \(^{2}\) and the coefficient of kinetic friction between the drum and the brake shoe is \(0.40 .\) Knowing that the initial angular velocity is 240 rpm clockwise, determine the force that must be exerted by the hydraulic cylinder if the system is to stop in 75 revolutions.
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